Mistake in Bartle's proof of Hake's Theorem? Here is Bartle's proof of Hake's Theorem found in "A Modern Theory of Integration". I think there is a mistake in the highlighted line:
The Theorem: $f:[a,b]\to \mathbb{R}$ is gauge integrable if and only if it is gauge integrable in $[a,x]\ \forall x\in [a,b)$ and the limit $\lim_{x\to b}\int_a^xf$ is finite. In that case
\begin{equation}\int_a^bf=\lim_{x\to b^-}\int_a^xf\end{equation}
The problem is in the $(\impliedby)$ proof:
Let $(c_n)$ be a strictly increasing sequence with $c_0=a$, $c_n\to b$. As $\lim_{n\to +\infty}\int_{c_0}^{c_n}f=\lim_{x\to b^-}\int_a^xf=L\in \mathbb{R}$ and $c_n\to b$, $\exists r\in \mathbb{N}$ so that
\begin{equation}\forall x\in (c_r,b)\ \left|\int_a^xf-L\right|<\epsilon\text{ and }\left|f(b)\right|(b-c_r)<\epsilon\end{equation}
By the integrability in $[c_{n-1},c_n]$, there exists a gauge $\delta_n$ on $[c_{n-1},c_n]$ so that if $\dot{\mathcal{P}}_n<<\delta_n$,
\begin{equation}\left|S(f,\dot{\mathcal{P}}_n)-\int_{c_{n-1}}^{c_n}f\right|<\frac{\epsilon}{2^n}\end{equation}
Without loss of generality,
\begin{align*}\delta_1(c_0)&\le \frac12(c_1-c_0)\tag{i}\\
\delta_{n+1}(c_n)&\le \min\left\{\delta_n(c_n),\frac12(c_n-c_{n-1}),\frac12(c_{n+1}-c_n)\right\}\tag{ii}\\
\delta_n(x)&\le \min\left\{\frac12(x-c_{n-1}),\frac12(c_n-x)\right\}\text{ for }x\in (c_{n-1},c_n)\tag{iii}
\end{align*}
We define a gauge in $[a,b]$ by 
\begin{equation}\delta(x)=\begin{cases}\delta_n(x)&\text{ if, }x\in [c_n,c_{n+1})\\
b-c_N&\text{ if, }x=b\end{cases}\end{equation}
Let $\mathcal{P}=\left\{a=x_0<...<x_m=b\right\}$ partition $[a,b]$ with tags $t_i$ so that $\dot{\mathcal{P}}<<\delta$. 
As $b\not\in\cup_{n=0}^{\infty}[c_n,c_{n+1})=[a,b)$, this forces $b$ to be a tag of $[x_{m-1},b]$. But then, as $t_m=b$,
\begin{equation}c_r=b-\delta(b)<x_{m-1}\end{equation}
Let $s\in \mathbb{N}$ be the smallest integer with $x_{m-1}\le c_s$ so that $r\le s$.
If $k=1,...,s-1$ then condition (iii) implies that the point $c_k$ must be a tag of the sub-interval in $\mathcal{P}$ that contains it. ...
My argument: Suppose $c_k\in [x_p,x_{p+1}]$. But, $[x_p,x_{p+1}]$ may also contain other points of the sequence (finitely many). Let $c_{q},c_{q+1},...,c_k,c_{k+1},...,c_l$ be all these points. Suppose $t_p$ is the tag of $[x_p,x_{p+1}]$. If $t_p\neq c_q,c_{q+1},...,c_l$
and $t_p>c_k$ then $t_p\in (c_i,c_{i+1})$ with $c_i\ge c_k$. Condition (iii) and the fact that $\dot{\mathcal{P}}<<\delta$ imply that 
$$c_k> t_p-\delta(t_p)=t_p-\delta_i(t_p)\ge t_p-\frac{t_p-c_i}2=\frac{t_p+c_i}2\ge \frac{c_k+c_k}2=c_k$$
which is a contradiction. Similarly if $t_p<c_k$. What I have shown is that the tag must be a point of the sequence. But why must it be $c_k$ like Bartle says?. This fact is crucial to the proof. Bartle then continues:
Using the right-left procedure we may assume $c_0,c_1,...,c_{s-1}\in \mathcal{P}$.
That can't be done (I think). Indeed consider the simple case where the only points of the sequence in 
$[x_p,x_{p+1}]$ are $c_k$ and $c_{k+1}$. The tag $t_p$ must be one of $c_k,c_{k+1}$. Suppose it were $c_k$. Then $c_k\in \mathcal{P}$ and $c_{k+1}\in (c_k,x_{p+1})$. How is it then possible to prove $c_{k+1}$ is the tag of $[c_k,x_{p+1}]$?. $c_k$ seems like an equally strong candidate for this.
EDIT: I also think there is a gap in "As $b\not\in\cup_{n=0}^{\infty}[c_n,c_{n+1})=[a,b)$, this forces $b$ to be a tag of $[x_{m-1},b]$." Why can't a point $c_n$ be the tag?
 A: I don't have the book, so I can only go on what you've quoted.  But based on that, the indexing seems off in your definition of $\delta$: the gauge $\delta_n$ is defined on $[c_{n-1},c_n]$, so we should have $\delta(x) = \delta_{n+1}(x)$ if $x\in [c_n,c_{n+1})$, not $\delta_n(x)$.  Maybe this is a Bartle's mistake, or maybe you made a typo in copying it.
Anyway, with this correction, I think that condition (ii) solves the problems.  Condition (ii) guarantees that $\delta(c_n)$ is smaller than the distance from $c_n$ to any other point of the sequence.  Thus, a sub-interval of which $c_n$ is the tag cannot contain any other $c_k$.  Similarly, it cannot contain $b$, since $b$ is greater than all the $c_k$'s.
A: Ok, the fact that if there is almost a point $c_{k}$ of the sequence in $[x_{k-1},x_{k}]$ then one of them must be the tag seems to be true but in addition we have the following:
By the definition of $\delta$: $\delta_{1}(c_{0})\leq \frac{1}{2}(c_{1} - c_{0})$ $\;$. Suppose that there are two points $c_{0}$ and $c_{1}$ of the sequence. One of them must be the tag , let $c_{0}$ be the tag. For the $\delta$-fine condition:
$$
[x_{k-1},x_{k}] \subseteq [t_{k}-\delta(c_{0}) , t_{k}+\delta(c_{0})]
$$
Therefore
$$
| x_{k}- x_{k-1} | \leq 2 \delta(c_{0}) \leq |c_{1} - c_{0}|
$$
which is a contradiction except when $c_{0}=x_{k-1}$ and $c_{1}=x_{k}$ in which case we use the left-right procedure to obtain both subintervals that have $c_{0}$ and $c_{1}$ as tags. 
Conclusion: There cannot be more than one element c_{k} of the sequence in each subinterval $[x_{k-1},x_{k}]$ containing $c_{k}$ and this $c_{k}$ must be the tag.
A: As @MikeShulman mentioned, there is a typo by the OP. An example of gauge $\delta$ satisfying the condition (i)-(iii) is
\begin{align}
\delta(x)=\left\{\begin{array}[lcr]
\min\Big(\delta_n(x),\frac{x-c_{n-1}}{2},\frac{c_n-x}{2}\Big) &\text{if} & c_{n-1}<x<c_n\\
\min\Big(\delta_1(c_0),\frac{c_1-c_0}{2}\Big) &\text{if}& x=c_0=a\\
\min\Big(\delta_n(c_n),\delta_{n+1}(c_n),\frac{c_n-c_{n-1}}{2},\frac{c_{n+1}-c_n}{2}\Big) &\text{if} & x=x_n, n\geq1\\
b-c_r &\text{if} & x=b
\end{array}
\right.
\end{align}
